Question

The equation of a common tangent to the parabolas $y=x^2$ and $y=-(x-2{)}^2$ is

• A. $y=4(x-2)$
• B. $y=4(x-1)$
• C. $y=4(x+1)$
• D. $y=4(x+2)$

Answer 1

Answer: (B)

Equation of tangent of $y=x^2$ be
$tx=y+a{t}^2......$(1)
$y=tx-\frac{t^2}{4}$
Solve with $y=-(x-2{)}^2$
$tx-\frac{t^2}{4}=-(x-2{)}^2$
$x^2+x(t-4)-\frac{t^2}{4}+4=0$
$D=0$
$(t-4{)}^2-4\cdot \left(4-\frac{t^2}{4}\right)=0$
$t^2-4t=0$
$t=0$ or $t=4$
From eq. (1), required common tangent is $y=4(x-1)$